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3 years ago

2. What mechanism would you use to safely capture emissions?

Physics

1 answer:

devlian [24]3 years ago

5 0

Carbon capture is the safest way to capture emission.

<u>Explanation:</u>

Emission of green house gases is one of the most lurking threat above humans. So one of the best ways to reduce emission is to capture the gases like carbon di-oxide or carbon monoxide, nitrous oxide etc. So mostly, the carbon di-oxide gas which is produced as waste is captured using adsorption, absorption, chemical looping etc.

In this method, they are captured from high source points like cement factory etc and then stored in a secluded region. Thus, the adsorption of these gases and then storing them in secluded region is the best option for capturing emission.

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Hey can anyone help me with my physics exam

Alona [7]

Answer:

25 is the correct answer

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2 years ago

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Suppose you are observing the interference pattern formed by a Michelson interferometer in a laboratory and a joking colleague h

Troyanec [42]

Answer:

Explanation:

In Michelson interferometer , two light waves from different directions are made to overlap so that fringes are formed on the screen due to interference . In it, two monochromatic and coherent light are made to overlap which have some path difference or phase difference. They form dark and bright fringes .

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3 years ago

The energy of the electron in Hydrogen atom can be shown to be given by En = -13.6/n^2 eV, where n represents the principal quan

Tpy6a [65]

Answer:

This represents radiation in ultra-violet region .

Explanation:

Energy of the orbit where n = 3 is given as follows

$E_3 =- \frac{13.6 }{3^2}$

$E_3 =- \frac{13.6 }{9}$ = -1.511 eV

Energy of the orbit where n = 1 is given as follows

$E_1 =- \frac{13.6 }{1^2}$

$E_1 =- \frac{13.6 }{1}$ = 13.6 eV

Difference of [tex]E_3 and [tex]E_1 = - 1.511+ 13.6

= 12.089 eV.

The wavelength of light having this energy in nm is given by the expression as follows

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3 years ago

The cockroach Periplaneta americana can detect a static electric field of magnitude 8.50 kN/C using their long antennae. If the

erica [24]

Answer:

0.235 nC

Explanation:

Given:

$E$ = the magnitude of electric field = $8.50\ kN/C =8.50\times 10^{3}\ N/C$

$F$ = the magnitude of electric force on each antenna = $2.00\ \mu N =2.00\times 10^{-6}\ N$

$q$ = The magnitude of charge on each antenna

Since the electric field is the electric force applied on a charged body of unit charge.

$\therefore E = \dfrac{F}{q}\\\Rightarrow q =\dfrac{F}{E}\\\Rightarrow q =\dfrac{2.00\times 10^{-6}\ N}{8.50\times 10^{3}\ N/C}\\\Rightarrow q =0.235\times 10^{-9}\ C\\\Rightarrow q =0.235\ nC$

Hence, the value of q is 0.235 nC.

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3 years ago

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Answer:

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